Why Are My Logarithmic Equation Solutions Extraneous?

[cbarker1]

Solve for x

$\log\left({2-x}\right)+\log\left({3-x}\right)=\log\left({12}\right)$
$\log\left({(2-x)(3-x})\right)=\log\left({12}\right)$
$\left(2-x)\right)\left(3-x)\right)=12$
${x}^{2}-5x-6=0$
$\left(x-2)\right)\left(x-3)\right)=0$
$x=2, x=3$

I have a problem with solutions because both is extraneous.

$\log\left({2-2}\right)+\log\left({3-2}\right)=\log\left({12}\right)$
$\log\left({0}\right)+\log\left({1}\right)=\log\left({12}\right)$
This solution is extraneous as well as the other solution.

The solution in the back of the book is -1.

Where did I made a mistake?Thank you

 

[earboth]

Solve for x

$\log\left({2-x}\right)+\log\left({3-x}\right)=\log\left({12}\right)$
$\log\left({(2-x)(3-x})\right)=\log\left({12}\right)$
$\left(2-x)\right)\left(3-x)\right)=12$
${x}^{2}-5x-6=0$
$\color{red}(x-2)(x-3)=0$
$x=2, x=3$

...

Hello,

I've marked the line where you made a tiny (but fatal) mistake.

After facvtorization you should come out with

\(\displaystyle x^2-5x-6 = (x+1)(x-6)=0\)